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IB Biology B3.1: Gas exchange worked solutions

Worked solutions to every practice question in B3.1: Gas exchange of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
In which process are oxygen used and carbon dioxide produced inside the cells of the body? (A) Ventilation (B) Gas exchange at the alveolus (C) Cell respiration (D) Transpiration
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Cell respiration uses oxygen and releases carbon dioxide inside cells. Ventilation moves air, gas exchange is diffusion across a surface, and transpiration is water loss from leaves.

Final answer

C

Question B1

[1 mark]
Which feature of an alveolus increases the rate of gas exchange? (A) A wall several cells thick (B) A wall one cell thick (C) A dry inner lining (D) Few surrounding capillaries
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A wall one cell thick makes the diffusion distance very short. The other options lengthen the path, slow dissolving of gases or weaken the gradient.

Final answer

B

Question C1

[1 mark]
Which row describes inhalation? (A) diaphragm contracts; external intercostals relax; pressure falls (B) diaphragm contracts; external intercostals contract; pressure falls (C) diaphragm relaxes; external intercostals contract; pressure rises (D) diaphragm relaxes; external intercostals relax; pressure rises
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In inhalation the diaphragm and external intercostals both contract, the thorax volume rises and the pressure in the lungs falls below atmospheric pressure.

Final answer

B

Question D1

[1 mark]
Which muscles contract during forced exhalation but not during inhalation? (A) Diaphragm and external intercostals (B) Internal intercostals and abdominal muscles (C) Diaphragm and abdominal muscles (D) External and internal intercostals
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Internal intercostals pull the ribs down and in and abdominal muscles push the diaphragm up. The diaphragm and external intercostals contract in inhalation.

Final answer

B

Question F1

[1 mark]
In a transverse section of a leaf, which tissue lies directly below the upper epidermis and holds the most chloroplasts? (A) Spongy mesophyll (B) Palisade mesophyll (C) Lower epidermis (D) Xylem
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The tightly packed palisade cells lie just below the upper epidermis, where most light arrives, and are full of chloroplasts.

Final answer

B

Question G1

[1 mark]
A plant is moved from dry air into humid air. What happens to the rate of transpiration, and why? (A) It rises, because the vapour gradient is steeper. (B) It falls, because the vapour gradient is shallower. (C) It falls, because the stomata open wider. (D) It is unchanged, because the stomata are the same.
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Humid air already holds water vapour, so the gradient from the moist air spaces to the air outside is shallower and less vapour leaves.

Final answer

B

Question H1

[1 mark]
Oxygen diffuses from air in an alveolus, through an alveolar cell, then through a capillary endothelial cell and into the plasma. How many cell membranes does it cross? (A) 2 (B) 3 (C) 4 (D) 5
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It enters and leaves each of the two cells it passes through, so 2+2=4 membranes. It has not yet entered a red blood cell.

Final answer

C

Question I1

[2 marks]
A field of view of 0.20 mm \times 0.20 mm of a lower epidermis contains 9 stomata. (a) Calculate the area of the field in mm^2. [1] (b) Calculate the stomatal density. [1]
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(a) 0.20\times0.20=0.04\,\mathrm{mm^2} (1). (b) 9\div0.04=225\,\mathrm{mm^{-2}} (1).

Final answer

(a) 0.04 mm^2. (b) 225 mm^{-2}.

Question J1

HL[1 mark]
What is meant by cooperative binding of oxygen to haemoglobin? (A) Each haem group binds oxygen independently of the others. (B) Binding of one oxygen molecule makes it easier for the next to bind. (C) Oxygen and carbon dioxide bind at the same site. (D) Oxygen binds only when the pH is low.
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The first oxygen changes the shape of the molecule so that the remaining haem groups bind oxygen more easily, which gives the S-shaped curve.

Final answer

B

Tutorial practice

Question K2

[4 marks]
Each alveolus has a wall one cell thick, is lined with a water film containing surfactant and is surrounded by capillaries. (a) Explain the role of surfactant. [2] (b) Explain how continuous blood flow through the capillaries keeps the oxygen gradient steep. [2]
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(a) Surfactant reduces the surface tension of the water film (1), which stops the alveoli collapsing and sticking together, so the surface area is kept (1). (b) Blood flowing past removes the oxygen that has just diffused in (1), and replaces it with blood low in oxygen, so the concentration difference stays high (1).

Final answer

(a) It lowers surface tension of the water film, so alveoli do not collapse or stick, keeping the area. (b) Blood carries oxygen away and brings blood low in oxygen, so the difference stays large.

Question L2

[4 marks]
A student breathes in deeply. (a) Explain how contraction of the diaphragm and the external intercostal muscles brings air into the lungs. [4]
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(a) The diaphragm contracts and flattens (1) and the external intercostals raise the ribs up and out (1). The volume of the thorax increases (1), so the pressure in the lungs falls below atmospheric pressure and air flows in down the pressure gradient (1).

Final answer

(a) Diaphragm flattens, ribs up and out, thorax volume rises, pressure falls below atmospheric, air flows in.

Question M2

[4 marks]
A student breathed into a spirometer. Readings from the trace: top of a normal breath 3.1 dm^3, bottom of a normal breath 2.6 dm^3; top of the deepest breath 6.2 dm^3, bottom of the deepest breath out 1.3 dm^3; 5 normal breaths in 20 s. (a) Calculate the tidal volume. [1] (b) Calculate the vital capacity. [1] (c) Calculate the ventilation per minute at rest. [2]
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(a) 3.1-2.6=0.5\,\mathrm{dm^3} (1). (b) 6.2-1.3=4.9\,\mathrm{dm^3} (1). (c) 5 breaths in 20 s is 5\times3=15 breaths per minute (1); 0.5\times15=7.5\,\mathrm{dm^3\,min^{-1}} (1).

Final answer

(a) 0.5 dm^3. (b) 4.9 dm^3. (c) 7.5 dm^3 min^{-1}.

Question N2

[4 marks]
A student measures vital capacity with a spirometer. (a) Outline a method the student could follow to obtain a reliable value. [4]
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Any four. Calibrate the spirometer or check it reads zero (1). Use soda lime to absorb carbon dioxide and a clean mouthpiece with a nose clip so there is no leak (1). The person breathes normally, then breathes in as deeply as possible and out as fully as possible (1). Read the volume from the trace (1). Repeat at least three times and calculate a mean (1). Same posture and rest period each time (1).

Final answer

Calibrate; soda lime and clean mouthpiece, nose clip, no leaks; breathe normally, then deepest breath in and out as far as possible; read volume from trace; repeat three times and average.

Question O2

[4 marks]
In a transverse section of a dicotyledonous leaf the layers from the upper surface are: a thin transparent layer with a waxy coating (P); a layer of tall cells packed tightly together (Q); a layer of rounded cells with large spaces between them (R); and a thin layer with small pores, each bounded by a pair of cells (S). (a) Identify layers P, Q, R and S. [2] (b) Explain how the structure of Q suits photosynthesis. [2]
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(a) P upper epidermis and cuticle, Q palisade mesophyll, R spongy mesophyll, S lower epidermis with stomata: two or three correct (1); all four correct (1). (b) The cells are full of chloroplasts (1) and packed close to the upper surface, where most light enters, so they capture much light energy (1).

Final answer

(a) P epidermis with cuticle; Q palisade mesophyll; R spongy mesophyll; S lower epidermis with stomata. (b) Many chloroplasts and close packing near the light capture more light.

Question P2

[5 marks]
A leaf takes in carbon dioxide for photosynthesis in bright light. (a) Describe the route of a carbon dioxide molecule from the air outside to a chloroplast. [3] (b) Explain how the stomata open in light. [2]
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(a) The molecule enters through a stoma (1), diffuses through the air spaces between the mesophyll cells (1), dissolves in the moist wall of a mesophyll cell and diffuses into the cell and the chloroplast (1). (b) In light ions such as potassium enter the guard cells (1); water follows by osmosis, the guard cells become turgid and bow apart, opening the pore (1).

Final answer

(a) Stoma, air spaces, dissolves in water on mesophyll cell walls, diffuses into the cell and chloroplast. (b) Ions enter guard cells, water follows by osmosis, turgid cells bow apart.

Question R2

[4 marks]
A student uses a bubble potometer to compare water uptake of a leafy shoot with the lower leaf surfaces coated in petroleum jelly and with no coating. (a) Identify two variables that must be controlled. [2] (b) Predict the effect of the coating and explain it. [2]
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(a) Any two of: light intensity (1); temperature (1); air movement (1); humidity (1); leaf area or shoot size (1). (b) Uptake falls (1) because most stomata are on the lower surface and the coating blocks them, so less water vapour escapes (1).

Final answer

(a) Any two of light, temperature, wind, humidity, leaf area. (b) Uptake falls, because the coating blocks most stomata.

Question T2

HL[5 marks]
The muscles of a marathon runner produce large amounts of carbon dioxide. (a) Explain how this affects the oxygen dissociation curve of haemoglobin in the muscle capillaries. [3] (b) Outline why this is an advantage to the runner. [2]
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(a) Carbon dioxide and the hydrogen ions it forms bind to haemoglobin at sites other than the haem groups (1). This allosteric binding changes the shape of haemoglobin and lowers its affinity for oxygen (1), so the curve shifts to the right, a Bohr shift (1). (b) At the same partial pressure of oxygen, haemoglobin is less saturated, so more oxygen is released (1) to the actively respiring muscle, where the need is greatest (1).

Final answer

(a) More CO_2 and H^+ bind allosterically, lowering affinity; curve shifts right. (b) More oxygen is released at the same partial pressure, where it is needed.

Question Z3

[5 marks]
A student studies a transverse section of a leaf and counts stomata on the lower surface in five fields of view, each 0.20 mm \times 0.20 mm: 12, 9, 11, 14 and 9. (a) State two features of a plan diagram of the section. [2] (b) Calculate the mean stomatal density. [2] (c) Suggest one way to make the estimate more reliable. [1]
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(a) Any two: shows outlines of tissues and no individual cells (1); tissues drawn in correct proportion (1); every tissue labelled (1). (b) Mean count \frac{12+9+11+14+9}{5}=11 (1); 11\div0.04=275\,\mathrm{mm^{-2}} (1). (c) Count more fields on more leaves, and on both surfaces (1).

Final answer

(a) Tissue outlines, not cells; correct proportions; labels. (b) 275 mm^{-2}. (c) Count more fields and more leaves.

Advanced practice

Question U3

HL[5 marks]
The table gives the percentage saturation of adult and foetal haemoglobin at different partial pressures of oxygen. Oxygen (kPa) 2, 3, 4, 6, 8. Adult saturation (%) 18, 40, 59, 81, 90. Foetal saturation (%) 35, 62, 78, 91, 96. (a) Compare the two sets of data. [2] (b) Calculate the difference in saturation at 3 kPa, in percentage points. [1] (c) Explain how foetal haemoglobin obtains oxygen in the placenta. [2]
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(a) Foetal haemoglobin is more saturated at every pressure (1); both curves rise steeply and then level off towards saturation (1). (b) 62-40=22 percentage points (1). (c) Foetal haemoglobin has a higher affinity for oxygen (1), so at the low partial pressure in the placenta it becomes saturated by taking oxygen from maternal haemoglobin, which releases it (1).

Final answer

(a) Foetal is higher at every pressure; both rise and level off. (b) 22 percentage points. (c) Higher affinity, so it takes oxygen from maternal haemoglobin at low pressure.

Question V3

[5 marks]
Two species were grown at two carbon dioxide levels. The table gives the mean stomatal density of ten leaves with one standard deviation. A student claims that higher carbon dioxide lowers stomatal density in plants. Species A: 220 +- 18 at 400 ppm, 160 +- 15 at 800 ppm. Species B: 130 +- 12 at 400 ppm, 128 +- 14 at 800 ppm (mm-2). (a) Calculate the percentage change in density of species A. [1] (b) Using the standard deviations, suggest whether the difference for species B is likely to be real. [2] (c) Evaluate the student’s claim. [2]
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(a) \frac{160-220}{220}\times100=-27.3\% (1). (b) The means differ by only 2 mm^{-2}, much less than one standard deviation (12 and 14) (1), so the ranges overlap and the difference is probably not real (1). (c) Species A supports the claim but species B does not (1); only two species and two carbon dioxide levels were tested, so the claim cannot be generalised to all plants (1).

Final answer

(a) -27.3\%. (b) Difference of 2 is far smaller than the SDs (12, 14), so probably not real. (c) Supported for A, not B; two conditions, ten leaves; cannot generalise to all plants.

Question W3

[4 marks]
A model of a fish gill allows the flow of water over the filaments and the flow of blood through them to be switched on or off separately. The relative rate of oxygen uptake into the blood was recorded (values are synthetic). Both flowing 100; water stopped 28; blood stopped 22; both stopped 3 (relative uptake). (a) Explain why stopping the water flow reduces uptake. [2] (b) Deduce what the data show about the two flows. [2]
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(a) Without flow, the water next to the gill loses its oxygen to the blood (1), so the concentration difference between water and blood falls and diffusion slows (1). (b) Stopping either flow cuts uptake by more than 70% (1), so neither can replace the other and a gill needs both ventilation and blood flow to keep both gradients steep (1).

Final answer

(a) The water next to the gill loses oxygen, so the gradient falls. (b) Each single switch cuts uptake by over 70%; neither flow can replace the other.

Question Y3

[4 marks]
The table gives carbon dioxide uptake and water loss of a leaf at five stomatal apertures, in arbitrary units. The plant can afford a water loss of no more than 6 units. Aperture (micrometre) 0, 2, 4, 6, 8. CO2 uptake 0, 3.0, 5.0, 5.8, 6.0. Water loss 0.5, 2.0, 5.0, 9.0, 14.0 (arbitrary units). (a) Deduce the widest aperture that stays within the water budget. [1] (b) Calculate the uptake per unit of water lost at 2 and at 4 \mum. [2] (c) State the trade-off that the data show. [1]
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(a) At 4 \mum the water loss is 5, within 6, while at 6 \mum it is 9 (1). (b) \frac{3.0}{2.0}=1.5 (1); \frac{5.0}{5.0}=1.0 (1). (c) Opening the pore wider admits more carbon dioxide but loses disproportionately more water (1).

Final answer

(a) 4 \mum. (b) 1.5 at 2 \mum, 1.0 at 4 \mum. (c) More CO_2 means more water lost.

Connected questions

Question X1

Connected questions[6 marks]
B3.1 + C1.2 + B3.2. At rest a student breathes 15 times a minute with a tidal volume of 0.5 dm^3. During hard exercise the rate is 28 breaths per minute and the tidal volume 1.8 dm^3. (a) Calculate the ventilation per minute during exercise. [1] (b) Explain why exercise increases the muscles’ demand for oxygen. [2] (c) Explain how the lungs and the heart together increase oxygen delivery during exercise. [3]
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(a) 1.8\times28=50.4\,\mathrm{dm^3\,min^{-1}} (1). (b) Contracting muscle needs much more ATP (1), which is made by aerobic cell respiration in the mitochondria, a process that uses oxygen (1). (c) Faster and deeper ventilation renews the air in the alveoli more quickly (1); a faster heart rate increases blood flow through the lung capillaries and to the muscles (1); together they keep the oxygen concentration gradient steep and carry more oxygen to the muscles (1).

Final answer

(a) 50.4 dm^3 min^{-1}. (b) Muscle contraction needs more ATP, made by aerobic respiration in mitochondria, which uses oxygen. (c) Faster ventilation renews alveolar air; faster heart rate speeds blood flow; both keep the gradient steep.

Question X2

Connected questions[6 marks]
B3.1 + D2.3 + B3.2. On a warm dry day a tree loses water vapour from its leaves. A shoot of the same tree is placed in a humid bag. (a) State why the loss of water vapour cannot be avoided when the stomata are open. [1] (b) Explain how loss of water from the leaf draws water up the xylem. [3] (c) Predict and explain the effect of the humid bag on the movement of water in the shoot. [2]
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(a) The pores that admit carbon dioxide also let water vapour diffuse out (1). (b) Water evaporates from the mesophyll cell walls, lowering the water potential there (1) so that water moves from the xylem into the leaf along the water potential gradient (1); cohesion between water molecules pulls the whole column up the xylem (1). (c) Water movement slows (1) because the humid air reduces the vapour gradient, so less water is lost and less is pulled up (1).

Final answer

(a) The stomata that admit CO_2 also let vapour out. (b) Evaporation lowers water potential in the leaf; water moves in from the xylem by osmosis; cohesion pulls the column up. (c) Slows, because the vapour gradient is shallower.

Question X3

Connected questions[6 marks]
B3.1 + C1.3 + D4.3. Stored leaves of one species show a mean stomatal density of 180 mm^{-2} in the year 1900 and 144 mm^{-2} in the year 2000. Atmospheric carbon dioxide rose over the same period. (a) Calculate the percentage change in stomatal density. [1] (b) Explain how a lower stomatal density affects carbon dioxide uptake and water loss. [2] (c) Evaluate whether these data prove that rising carbon dioxide caused the change. [3]
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(a) \frac{144-180}{180}\times100=-20.0\% (1). (b) Fewer pores let in less carbon dioxide (1) but also lose less water vapour (1). (c) The data show a negative correlation, not cause (1); only one species and two dates were sampled (1), and temperature, water supply or light could also have changed stomatal density (1).

Final answer

(a) -20.0\%. (b) Fewer pores admit less CO_2 but lose less water. (c) Correlation only; one species, two dates; other factors such as temperature or water supply.

Question X4

Connected questions · HL[6 marks]
B3.1 + D4.1 + B3.2. The oxygen dissociation curve of a mammal that lives at high altitude lies to the left of the curve of a related lowland species. (a) State what the position of the curve shows about the affinity of its haemoglobin for oxygen. [1] (b) Explain why this is an advantage at high altitude. [2] (c) Explain how natural selection could have produced this difference. [3]
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(a) The haemoglobin has a higher affinity for oxygen (1). (b) The partial pressure of oxygen in the air is low at altitude (1), so haemoglobin with a higher affinity still becomes well saturated in the lungs (1). (c) The population had variation in haemoglobin genes (1); individuals with higher-affinity haemoglobin survived and reproduced more (1), passing the allele on, so its frequency rose over generations (1).

Final answer

(a) Higher affinity. (b) At low partial pressure of oxygen it still loads fully in the lungs. (c) Variation in haemoglobin; high-affinity individuals survive and breed more; allele frequency rises.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Gas exchange from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book

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